Solve in $\mathbb R\quad 5^\sqrt{x} - 5^{x-7} = 100$

My sugestion. Proceed as suggested by Dr. Sonnhard Graubner. Set the function $F:(0,10)\to \mathbb{R}$ by $$ F(x)=5^x-\frac{1}{5^7}5^{x^2}-100. $$ In the absence of a method of finite steps to solve an equation $F(x)=0$ there is a powerful method of resolution by interaction. The Kantorovich's theorem on Newton's interactions. The method sometimes (very rarely) results in a finite step method and thus exact solution. Use the Kantorovich's theorem on Newton's method in its classical formulation.

Let $I\subseteq \mathbb{R}$ and $F:{I}\to \mathbb{R}$ a continuous function, continuously differentiable on $\mathrm{int}(I)$. Take $x_0\in \mathrm{int}(I)$, $L,\, b>0$ and suppose that

.1 $F '(x_0)$ is non-singular,

.2 $ \| F'(x_0)^{-1}\left[ F'(y)-F'(x)\right] \| \leq L\|x-y\| \;\;$ for any $x,y\in I$,

.3$ \|F'(x_0)^{-1}F(x_0)\|\leq b$,

.4 $2bL\leq 1$.

Define \begin{equation} t_*:=\frac{1-\sqrt{1-2bL}}{L},\qquad t_{**}:=\frac{1+\sqrt{1-2bL}}{L}. \end{equation} If $ [x_0-t_*,x_0+t_*]\subset I, $ then the sequences $\{x_k\}$ generated by Newton's Method for solving $F(x)=0$ with starting point $x_0$, \begin{equation} \label{ns.KT} x_{k+1} ={x_k}-F'(x_k) ^{-1}F(x_k), \qquad k=0,1,\cdots, \end{equation} is well defined, is contained in $(x_0-t_*,x_0+t_*)$, converges to a point $x_*\in [x_0-t_*,x_0+t_*]$ which is the unique zero of $F$ in $[x_0-t_*,x_0+t_*]$ and \begin{equation} \label{eq:q.conv.x} \|x_*-x_{k+1}\|\leq \frac{1}{2} \|x_*-x_k \|, \qquad k=0,1,\,\cdots. \end{equation} Moreover, if assumption .4 holds as an strict inequality, i.e. $2bL<1$, then \begin{equation} \|x_*-x_{k+1}\|\leq\frac{1-\theta^{2^k}}{1+\theta^{2^k}} \frac{ L}{2\sqrt{1-2bL}}\|x_*-x_k\|^2\leq \frac{ L}{2\sqrt{1-2bL}}\|x_*-x_k\|^2, \quad k=0,1,\cdots, \end{equation} where $\theta:=t_*/t_{**}<1$, and $x_*$ is the unique zero of $F$ in $[x_0-t_*,x_0+t_*]$ for any $\rho$ such that $ t_*\leq\rho<t_{**},\qquad [x_0-\rho,x_0+\rho]\subset I.$